【Leetcode】 748. Shortest Completion

Given a string licensePlate and an array of strings words, find the shortest completing word in words.

A completing word is a word that contains all the letters in licensePlate. Ignore numbers and spaces in licensePlate, and treat letters as case insensitive. If a letter appears more than once in licensePlate, then it must appear in the word the same number of times or more.

For example, if licensePlate = "aBc 12c", then it contains letters 'a', 'b' (ignoring case), and 'c' twice. Possible completing words are "abccdef", "caaacab", and "cbca".

Return the shortest completing word in words. It is guaranteed an answer exists. If there are multiple shortest completing words, return the first one that occurs in words.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step","steps","stripe","stepple"]
Output: "steps"
Explanation: 
licensePlate contains letters 's', 'p', 's' (ignoring case), and 't'.
"step" contains 't' and 'p', but only contains 1 's'.
"steps" contains 't', 'p', and both 's' characters.
"stripe" is missing an 's'.
"stepple" is missing an 's'.
Since "steps" is the only word containing all the letters, that is the answer.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks","pest","stew","show"]
Output: "pest"
Explanation: 
licensePlate only contains the letter 's'. 
All the words contain 's', but among these "pest", "stew", and "show" are shortest. The answer is "pest" because it is the word that appears earliest of the 3.

Constraints:

1 <= licensePlate.length <= 7
licensePlate contains digits, letters (uppercase or lowercase), or space ' '.
1 <= words.length <= 1000
1 <= words[i].length <= 15
words[i] consists of lower case English letters.

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AC:

/*
 * @lc app=leetcode.cn id=748 lang=cpp
 *
 * [748] 最短补全词
 */

// @lc code=start
class Solution {
    
    
public:
    string shortestCompletingWord(string licensePlate, vector<string>& words) {
    
    
        array<int, 26> cnt{
    
    };
        for(char ch : licensePlate)
        {
    
    
            if(isalpha(ch))
            {
    
    
                cnt[tolower(ch) - 'a']++;
            }
        }

        int idx = -1;
        for(int i = 0; i < words.size(); i++)
        {
    
    
            array<int, 26> c{
    
    };
            for(char ch : words[i])
            {
    
    
                c[ch - 'a']++;
            }

            bool ok = true;
            for(int j = 0; j < 26; j++)
            {
    
    
                if(c[j] < cnt[j])
                {
    
    
                    ok = false;
                    break;
                }
            }

            if(ok && (idx < 0 || words[i].length() < words[idx].length()))
            {
    
    
                idx = i;
            }
        }
        return words[idx];
    }
};
// @lc code=end

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After reading it for a long time, it exceeded my estimated time, and I still didn't understand the meaning of the title, so I just looked at the official analysis .
Suddenly realized!
Turns out, there are so few steps:

• Count licensePlatethe number of occurrences of letters (at the same time, use toslow()the function to convert uppercase letters to lowercase)
• Traverse word, record subscripts, and count the number of occurrences of each word in it
• The number of returns is not less than licensePlatethe number of times and in the case of equality, the first one is preferred.

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